NCERT New Syllabus
Mathematics

Class: 10

Exercise 1.1

Topic: Real
Numbers

**Understanding Real Numbers: Key to Mathematical Mastery.**

Welcome to the path of discovering one of the most fundamental topics in mathematics: real numbers. This chapter from the NCERT class 10 syllabus offers the framework for comprehending advanced mathematical concepts by concentrating on properties, theorems, and real-number operations. In this blog, we'll look at exercise 1.1, breaking down each difficulty carefully and addressing it. Whether you're studying for an exam or creating a solid foundation, this post will easily guide you through each answer. Let's discover the power of real numbers together!

1) Theorem 1.1 (Fundamental Theorem of Arithmetic):

Every composite number may be written as a unique product of prime numbers, regardless of how the prime elements are organized. This means that each composite number has a unique prime factorisation except the prime number sequence.

2) For
any two positive integers a and b, the relationship between their HCF and

LCM is given by: HCF (a, b) × LCM (a, b) = a × b.

3) This
formula allows us to easily calculate the LCM of two positive integers once

we know their HCF.

### EXERCISE 1.1

Q1. Express each number as a product of its prime factors:

(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

### Explanation:

Every composite number can be represented as a product of prime numbers.

### Solution:

**(i) 140**

1)

**Prime factor of 140: Note: Apply divisibility test of 2.**140 = 2 x 70 Note: Apply divisibility test of 2.

140 = 2 x 2 x 35 Note: Apply divisibility test of 5.

140 = 2 x 2 x 5 x 7

140 = 2^{2}x 5 x 7

2)

**The prime factorization of****140**is**2**^{2}**x 5 x 7**.**(ii) 156**

1)

**Prime factor of 156: Note: Apply divisibility test of 2.**156 = 2 x 78 Note: Apply divisibility test of 2.156 = 2 x 2 x 39 Note: Apply divisibility test of 3.

156 = 2 x 2 x 3 x 13

156 = 2^{2}x 3 x 13

2)

**The prime factorization of****156**is**2**.^{2}x 3 x 13**(iii) 3825**

1)

**Prime factor of 3825: Note: Apply divisibility test of 3.**3825 = 3 x 1275 Note: Apply divisibility test of 3.

3825 = 3 x 3 x 425 Note: Apply divisibility test of 5.

3825 = 3 x 3 x 5 x 85 Note: Apply divisibility test of 5.

3825 = 3 x 3 x 5 x 5 x 17

3825 = 3^{2}x 5^{2 }x 17

2)

**The prime factorization of****3825**is**3**.^{2}x 5^{2 }x 17**(iv) 5005**

1)

**Prime factor of 5005: Note: Apply divisibility test of 5.**5005 = 5 x 1001 Note: Apply divisibility test of 7.

5005 = 5 x 7 x 143 Note: Apply divisibility test of 11.

5005 = 5 x 7 x 11 x 13

2)

**The prime factorization of****5005**is**5 x 7 x 11 x 13**.**(v) 7429**

Note:

a) 2 is not a factor of 7429 as it is not an even number. So 4, 8, 16, and so on

are also not the factors of 7429.

b) 3 is not a factor of 7429 as the sum of the digits is 22, which is not divisible

by 3. So 6, 9, 12, 15, and so on are also not factors 7429.

c) 5 is not a factor of 7429 as the unit placed digit is not 0 or 5. So 5, 10, 15,

and so on are also not factors 7429.

d) 7 is not a factor of 7429. So 14, 21, and so on are also not factors 7429.

1)

**Prime factor of 7429: Note: Try directly dividing by 17.**7429 = 17 x 437 Note: Try directly dividing by 19.

7429 = 17 x 19 x 23

2)

**The prime factorization of****7429**is**17 x 19 x 23**.**Q2. Find the LCM and HCF of the following pairs of integers and verify that**

LCM × HCF = product of the two numbers.

(i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54

### Explanation:

1) To determine the LCM and HCF of given numbers, we first need to identify their

prime factors.

2) The HCF (Highest Common Factor) is found by multiplying the smallest powers

of the common prime factors between the numbers.

3) The LCM (Least Common Multiple) is calculated by multiplying the greatest

powers of each prime factor present in the numbers.

4) The relationship between HCF and LCM is given by the formula:

HCF (p, q) × LCM (p, q) = p × q, where p and q are any two numbers that are positive integers.

### Solution:

**(i) 26 and 91**

1) First, find the prime factors of 26:

26 = 2 x13

2) Next, find the prime factors of 91:

91 = 7 x13

3) The common factor between 26 and 91 is

**13**.4) Therefore, the

**HCF**of**26**and**91**is**13**.5) To get the LCM, multiply the individual prime factors:

LCM = 2 x 7 x13

LCM = 14 x13

LCM = 182.

6) Thus, the

**LCM**of**26**and**91**is**182**.7)

**Therefore,****LCM of 26 and 91 is 182, and HCF of 26 and 91 is 13.**8) We know 26 x 91 = 2366 -------- (equation 1).

9) Also, HCF x LCM = 13 x 182 = 2366 -------- (equation 2).

10) From Equation 1 and Equation 2, we confirm that:

HCF(26,91) x LCM(26,91) = 26 x 91.

**(ii) 510 and 92**

1) First, find the prime factors of 510:

510 =2x 255

510 =2x 3 x 85

510 =2x 3 x 5 x 17

2) Next, find the prime factors of 92:

92 =2x 4692 =2x 2 x 23

3) The common factor between 510 and 92 is

**2**.4) Therefore, the

**HCF**of**510**and**92**is**2**.5) To get the LCM, multiply the individual prime factors:

LCM =2x 2 x 3 x 5 x 17 x 23.

LCM = 4 x 15 x 17 x 23.

LCM = 23460.

6) Thus, the

**LCM**of**510**and**92**is**23460**.7)

**Therefore, the****LCM of 510 and 92 is 23460, and the HCF of 510 and 92 is 2.**8) We know 510 x 92 = 46920 -------- (equation 1)

9) Also, HCF x LCM = 2 x 23460 = 46920 -------- (equation 2).

10) From Equation 1 and Equation 2, we confirm that:

HCF(510,92) x LCM(510,92) = 510 x 92.

**(iii) 336 and 54**

1) First, find the prime factors of 336.

336 = 2 x 168336 = 2 x 2 x 84336 = 2 x 2 x 2 x 42336 = 2 x 2 x 2 x 2 x 21336 =2x 2 x 2 x 2 x3x 7

2) Next, find the prime factors of 54.

54 = 2 x 2754 = 2 x 3 x 954 =2x3x 3 x 3

3) The common factor between 336 and 54 is

**2 x 3**=**6**.4) Therefore, the

**HCF**of**336**and**54**is**6**.5) To get the LCM, multiply the individual prime factors:

LCM =2x 2 x 2 x 2 x3x 3 x 3 x 7

LCM = 16 x 27 x 7

LCM = 3024.

6) Thus, the

**LCM**of**336**and**54**is**3024**7)

**Therefore,****LCM of 336 and 54 is 3024, and HCF of 336 and 54 is 6.**8) We know 336 x 54 = 18144 -------- (equation 1).

9) Also, HCF x LCM = 6 x 3024 = 18144 -------- (equation 2).

10) From Equation 1 and Equation 2, we confirm that:

HCF(336, 54) x LCM(336, 54) = 336 x 54.

**Q3. Find the LCM and HCF of the following integers by applying the prime**

factorisation method.

**(i) 12, 15, and 21 (ii) 17, 23, and 29 (iii) 8, 9 and 25**

### Solution:

**(i) 12, 15 and 21**

1) Now we will find the prime factors of 12.

12 = 2 x 612 = 2 x 2 x3

2) Now we will find the prime factors of 15.

15 =3x 5

3) Now we will find the prime factors of 21.

21 =3x 7

4) Here, the common factor is

**3**.5) So

**HCF**of 12, 15, and 21 is**3**.6) Simmilarly, LCM = 2 x 2 x

**3**x 5 x 7 = 420.7)

**Thus, LCM is 420, and HFC is 3.**

**(ii) 17, 23 and 29**

1) Now we will find the prime factors of 17.

17 =1x 17

2) Now we will find the prime factors of 23.

23 =1x 23

3) Now we will find the prime factors of 29.

29 =1x 29

4) Here common factor is

**1**.5) The

**HCF**of 17, 23, and 29 is**1**.6) Simmilarly, LCM =

**1**x 17 x 23 x 29 = 11339.7)

**Thus, LCM is 11339, and HFC is 1.****(iii) 8, 9 and 25**

1) Now we will find the prime factors of 8.

8 = 1 x 2 x 48 =1x 2 x 2 x 2

2) Now we will find the prime factors of 9.

9 =1x 3 x 3

3) Now we will find the prime factors of 25.

25 =1x 5 x 5

4) Here common factor is

**1**.5) So

**HCF**of 8, 9, and 25 is**1**.6) Simmilarly, LCM =

**1**x 2 x 2 x 2 x 3 x 3 x 5 x 5 = 1800.7)

**Thus, LCM is 1800, and HFC is 1.****Q4. Given that HCF (306, 657) = 9, find LCM (306, 657).**

### Explanation:

1) We know that

**HCF (p, q) × LCM (p, q) = p × q**, where p and q are any two numbers that are positive integers.2) Out of the above 3 values, if 2 values are given, we can find the 3rd value.

### Solution:

1) Here HCF (306, 657) = 9.

2) So, HCF of 306 and 657 = 9.

3) Products of 306 and 657 are 306 x 657

4) We know that HCF (p, q) × LCM (p, q) = p × q, so,

[LCM (306, 657)] = [306 x 657] / HCF (306, 657)

[LCM (306, 657)] = [306 x 657] / 9

[LCM (306, 657)] = [34 x 657][LCM (306, 657)] = [22338]

5)

**Thus, LCM (306, 657) = 22338.****Q5. Check whether**

**6**

^{n}**can end with the digit 0 for any natural number n.**

### Explanation:

1) If any number ends with the digit 0, then that number must be divisible by 2 and 5 simultaneously.

2) To check whether 6

^{n}**ends with 0, we must find whether 6**^{n}**has 2 and 5 as prime factors or not.**### Solution:

1) Factors of 6

^{n}= (2 x 3)^{n}2) Here 2 is the factor of 6

^{n}, but 5 is not a factor of 6^{n}.3) So

**6**.^{n}can't end with 0 for any natural number n**Q6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.**

### Explanation:

1) Prime numbers have only 2 factors. 1 and the number itself. e.g. the prime number 5 has two factors. 1 and 5 itself.

2) The positive Number, which has factors other than 1, is known as a composite number.

### Solution:

1) The first expression is (7 × 11 × 13) + 13. We will simplify it.

(7 × 11 × 13 + 13) = 13 ((7 x 11 x 1) + 1)

= 13 (77 + 1)= 13 (78)= 13 (2 x 39)= 13 (2 x 3 x 13)

2) The first expression has factors as

**(2 x 3 x 13 x 13)**3) The number obtained from (7 × 6 × 5 × 4 × 3 × 2 × 1) + 5 is a composite number.

4) The second expression is (7 × 11 × 13) + 13. We will simplify it.

(7 × 6 × 5 × 4 × 3 × 2 × 1) + 5 = 5 ((7 x 6 x 4 x 3 x 2 x 1) + 1)

= 5 (1008 + 1)= 5 (1009)

5) The second expression has factors as

**(5 x 1009)**6) The number obtained from

**(7 × 6 × 5 × 4 × 3 × 2 × 1) + 5****is a composite**number.

**Q7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting point?**

### Explanation:

1) Sonia takes 18 minutes and Ravi takes 12 minutes to complete one round.

2) Now, we need to find out how many minutes will they meet again at the same point.

3) Here we will find LCM of 18 and 12 to get the time when both meet again at the starting point.

### Solution:

1) To find the LCM of 18 and 12, we first need to determine their prime factors.

2) Prime factorization of 18:

18 = 2 x 918 =2x3x 3

3) Prime factorization of 12:

12 = 2 x 612 =2x 2 x3

4) The common prime factor is 2 x 3 = 6.

5) The LCM (Least Common Multiple) is the product of all the highest powers of the

prime factors:

LCM =2x 2 x 3 x3= 36.

6) Therefore, the

**LCM**is**36**.7) After

**36 minutes Sonia and Ravi will meet again at the starting point**.

**Conclusion: Unveiling the Power of Real Numbers**

In this chapter on

*, we've explored the foundational blocks of mathematics, laying the groundwork for a deeper understanding of algebra and beyond. From the beauty of irrational numbers to the precision of the Euclidean algorithm, real numbers are at the core of every mathematical operation. As you continue to delve into the world of numbers, remember that these principles stretch far beyond the classroom, influencing technology, science, and everyday calculations. Keep exploring, keep calculating, and unlock the endless possibilities of real numbers!***Real Numbers**#RealNumbersUnveiled #Class10Math #NCERTSyllabus #NumberTheory #AlgebraEssentials #MathInLife #MathIsBeautiful #NCERTClass10 #Mathematics #NCERTMaths #Grade10Maths #MathSyllabus #NCERTSolutions #MathTips #LearnMath #MathConcepts #MathMadeEasy #RealNumberSystem #MathHelp #PrimeNumbers #MathForStudents #CBSEMath #MathEducation #MathLearning #simple method

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**NCERT New Syllabus Class 10 - Real Numbers Exercise 1.2**
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